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Pat will walk from intersection X to intersection Y along a route that

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Re: Pat will walk from intersection X to intersection Y along a route that  [#permalink]

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New post 19 Mar 2018, 09:58
Hi All,

There are a couple of ways that you can approach this question….

Since the answers are relatively small, there are at least 6 ways to get from X to Y, but no more than 16 ways to get from X to Y. In a pinch, you could draw pictures and physically find all of the possibilities.

If you're more interested in a "math" approach, you'll see that to get from X to Y you'll need to go 3 blocks "up" and 2 blocks "over" no matter how you get from X to Y.

Since you have to make 5 "moves" and 3 of them have to be "up", you have a Combination Formula situation….In other words…

5c3

5!/[3!2!] = 10

You COULD also say that to make 5 "moves" and 2 of them have to be "over", you could also use the combination formula in this way…

5c2

5!/[2!3!] = 10

It's the same answer because 5c3 is the same as 5c2.

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Re: Pat will walk from intersection X to intersection Y along a route that  [#permalink]

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New post 27 Mar 2018, 09:01
Pat will walk 3 blocks north and 2 blocks east to get to Y.

This can be represented as NNNEE

Total ways = 5!
Because she walks North 3 times and East 2 times, that has to be accounted for as 3!*2! to capture the minimum routes.

Minimum possible paths are now:

5!/ (3!*2!) = 10
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Re: Pat will walk from intersection X to intersection Y along a route that  [#permalink]

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New post 11 Dec 2019, 07:29
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haidzz wrote:
Image
Pat will walk from intersection X to intersection Y along a route that is confined to the square grid of four streets and three avenues shown in the map above. How many routes from X to Y can Pat take that have the minimum possible length?

A. Six
B. Eight
C. Ten
D. Fourteen
E. Sixteen

PS45461.01

Image
Attachment:
2019-09-21_1430.png


If we define Pat's path in a block-by-block manner, we can see that any route from X to Y will consist of 3 UPS and 2 RIGHTS.
So for example, if we let U represent walking one block UP, and let R represent walking one block RIGHT, one possible path is URURU.
Another possible path is UUURR
Another possible path is UURUR

So our question becomes, "In how many different ways can we arrange 3 U's and 2 R's?"

-----------ASIDE-----------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
---------------------------------

Now let's apply the MISSISSIPPI rule to arranging 3 U's and 2 R's
There are 5 letters in total
There are 3 identical U's
There are 2 identical R's
So, the total number of possible arrangements = 5!/[(3!)(2!)] = 10

Answer: C

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Re: Pat will walk from intersection X to intersection Y along a route that  [#permalink]

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New post 17 Dec 2019, 19:52
haidzz wrote:
Image
Pat will walk from intersection X to intersection Y along a route that is confined to the square grid of four streets and three avenues shown in the map above. How many routes from X to Y can Pat take that have the minimum possible length?

A. Six
B. Eight
C. Ten
D. Fourteen
E. Sixteen

PS45461.01

Image
Attachment:
2019-09-21_1430.png


Let V denote a step in the vertical direction and H denote a step in the horizontal direction. For instance, V-V-V-H-H denotes the path of walking along Avenue A until the intersection of 4th street and walking along 4th street until the point Y. Similarly, V-H-V-H-V denotes the path of walking along Avenue A, then walking along 2st street, then walking along Avenue B, then walking along 3rd street and, finally, walking along Avenue C to reach point Y.

We notice that a shortest path between point X and Y must include three V’s and two H’s. Further, any arrangement of three V’s and two H’s (i.e., any arrangement of the letters V-V-V-H-H) gives us a shortest path between X and Y. Using the permutations with indistinguishable objects formula, we see that there are 5! / (3!*2!) = (5 x 4)/2 = 10 such arrangements. Thus, there are 10 shortest paths between points X and Y.

Answer: C
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Re: Pat will walk from intersection X to intersection Y along a route that  [#permalink]

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New post 03 Jan 2020, 16:28
MBA HOUSE KEY CONCEPT: Combinatorics Permutation with repetition

5 steps = 2 to the right and 3 upward

Permutation of 5 with 2 and 3 repetitions = 5! / (2!)(3!) = 10

C :shocked
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New post 01 Feb 2020, 11:14
Bunuel wrote:
In order the length to be minimum Pat should only go UP and RIGHT: namely thrice UP and twice RIGHT.

So combination of UUURR: # of permutations of 5 letters out of which there are 3 identical U's and 2 identical R's is 5!/3!2!=10.

Answer: C.

If there were 5 streets and 4 avenues then the answer would be combination of UUUURRR: # of permutations of 7 letters out of which there are 4 identical U's and 3 identical R's is 7!/4!3!=35.



Hi Bunuel GMATPrepNow ScottTargetTestPrep

Silly doubt but why do we use permutations/arrangements here if order doesnt matter?

If he goes URURU or UUURR, the order doesnt matter right?
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Re: Pat will walk from intersection X to intersection Y along a route that  [#permalink]

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New post 01 Feb 2020, 11:29
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Top Contributor
Kritisood wrote:
Bunuel wrote:
In order the length to be minimum Pat should only go UP and RIGHT: namely thrice UP and twice RIGHT.

So combination of UUURR: # of permutations of 5 letters out of which there are 3 identical U's and 2 identical R's is 5!/3!2!=10.

Answer: C.

If there were 5 streets and 4 avenues then the answer would be combination of UUUURRR: # of permutations of 7 letters out of which there are 4 identical U's and 3 identical R's is 7!/4!3!=35.



Hi Bunuel GMATPrepNow ScottTargetTestPrep

Silly doubt but why do we use permutations/arrangements here if order doesnt matter?

If he goes URURU or UUURR, the order doesnt matter right?


URURU and UUURR are considered different paths.
So, order does matter.
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Re: Pat will walk from intersection X to intersection Y along a route that  [#permalink]

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New post 06 Feb 2020, 11:03
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Kritisood wrote:
Bunuel wrote:
In order the length to be minimum Pat should only go UP and RIGHT: namely thrice UP and twice RIGHT.

So combination of UUURR: # of permutations of 5 letters out of which there are 3 identical U's and 2 identical R's is 5!/3!2!=10.

Answer: C.

If there were 5 streets and 4 avenues then the answer would be combination of UUUURRR: # of permutations of 7 letters out of which there are 4 identical U's and 3 identical R's is 7!/4!3!=35.



Hi Bunuel GMATPrepNow ScottTargetTestPrep

Silly doubt but why do we use permutations/arrangements here if order doesnt matter?

If he goes URURU or UUURR, the order doesnt matter right?


The question is asking for the number of routes from X to Y which has the minimum number of lengths, i.e. 5 step routes. The routes URURU and UUURR are considered to be different routes and thus, the order matters in this question. That's why we are using permutations (with indistinguishable objects).
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Re: Pat will walk from intersection X to intersection Y along a route that  [#permalink]

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New post 14 Mar 2020, 17:33
Just so I can understand this concept better and wording of this question, would the solution be different for maximum possible length instead of the minimum possible length?
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Re: Pat will walk from intersection X to intersection Y along a route that   [#permalink] 14 Mar 2020, 17:33

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